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Page 1: Vibration Chapter05
Page 2: Vibration Chapter05

Mechanical Vibrations

Fifth Edition in SI UnitsSingiresu S. Rao

Page 3: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units3

Chapter 5Two Degree Freedom Systems

5

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© 2011 Mechanical Vibrations Fifth Edition in SI Units4

Chapter Outline

5.1 Introduction5.2 Equations of Motion for Forced Vibration5.3 Free Vibration Analysis of an Undamped System5.4 Torsional System5.5 Coordinate Coupling and Principal Coordinates5.6 Forced-Vibration Analysis5.7 Semidefinite Systems 5.8 Self-Excitation and Stability Analysis5.9 Transfer-Function Approach5.10 Solutions Using Laplace Transform5.11 Solutions Using Frequency Transfer Functions

Page 5: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units5

5.1Introduction

5.1

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© 2011 Mechanical Vibrations Fifth Edition in SI Units6

5.1 Introduction

• Two-degree-of-freedom systems are defined as systems that require two independent coordinates to describe their motion.

• For simplicity, a two-degree-of-freedom model can be used as shown in the figure.

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© 2011 Mechanical Vibrations Fifth Edition in SI Units7

5.1 Introduction

• The general rule for the computation of the number of degrees of freedom can be stated as follows:

=

No. of degrees of freedom of the system

No. of masses in the system x no. of

possible types of motion of each mass

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© 2011 Mechanical Vibrations Fifth Edition in SI Units8

5.1 Introduction

• As is evident from the systems shown in figures earlier on, the configuration of a system can be specified by a set of independent coordinates termed as generalized coordinates, such as length, angle, or some other physical parameters.

• Principle coordinates is defined as any set of coordinates that leads a coupled equation of motion to an uncoupled system of equations.

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© 2011 Mechanical Vibrations Fifth Edition in SI Units9

5.2Equations of Motion for Forced Vibration

5.2

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© 2011 Mechanical Vibrations Fifth Edition in SI Units10

5.2 Equations of Motion for Forced Vibration

• Consider a viscously damped two-degree-of-freedom spring-mass system, shown in the figure below

Page 11: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units11

5.2 Equations of Motion for Forced Vibration

• The application of Newton’s second law of motion to each of the masses gives the equations of motion:

• Both equations can be written in matrix form as

where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by

)2.5()()(

)1.5()()(

2232122321222

1221212212111

Fxkkxkxccxcxm

Fxkxkkxcxccxm

)3.5( )()(][)(][)(][ tFtxktxctxm

Page 12: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units12

5.2 Equations of Motion for Forced Vibration

• We have

• And the displacement and force vectors are given respectively:

• It can be seen that the matrices [m], [c], and [k] are symmetric:

where the superscript T denotes the transpose of the matrix.

322

221

322

221

2

1

][

][

0

0 ][

kkk

kkkk

ccc

cccc

m

mm

)(

)()(

)(

)()(

2

1

2

1

tF

tFtF

tx

txtx

][][],[][],[][ kkccmm TTT

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© 2011 Mechanical Vibrations Fifth Edition in SI Units13

5.3Free-Vibration Analysis of an Undamped System

5.3

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5.3 Free-Vibration Analysis of an Undamped System

• The solution of Eqs.(5.1) and (5.2) involves four constants of integration (two for each equation). We shall first consider the free vibration solution of Eqs.(5.1) and (5.2).

• By setting F1(t) = F2(t) = 0, and damping disregarded, i.e., c1 = c2 =

c3 = 0, and the equation of motion is reduced to:

• Assuming that it is possible to have harmonic motion of m1 and m2

at the same frequency ω and the same phase angle Φ, we take the solutions as

)5.5(0)()()()(

)4.5(0)()()()(

2321222

2212111

txkktxktxm

txktxkktxm

)6.5()cos()(

)cos()(

22

11

tXtx

tXtx

Page 15: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units15

5.3 Free-Vibration Analysis of an Undamped System

• Substituting into Eqs.(5.4) and (5.5),

• Since Eq.(5.7)must be satisfied for all values of the time t, the terms between brackets must be zero. Thus,

which represent two simultaneous homogenous algebraic equations

in the unknown X1 and X2.

)7.5( 0)cos()(

0)cos()(

2322

212

221212

1

tXkkmXk

tXkXkkm

)8.5(0)(

0)(

2322

212

221212

1

XkkmXk

XkXkkm

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5.3 Free-Vibration Analysis of an Undamped System

• For trivial solution, i.e., X1 = X2 = 0, there is no solution. For a

nontrivial solution, the determinant of the coefficients of X1 and X2

must be zero:

which is called the frequency or characteristic equation.

0

)(

)(det

212

12

2212

1

kkmk

kkkm

)9.5(0))((

)()()(223221

1322214

21

kkkkk

mkkmkkmm

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© 2011 Mechanical Vibrations Fifth Edition in SI Units17

5.3 Free-Vibration Analysis of an Undamped System

• Hence the roots are:

• The roots are called natural frequencies of the system.

)10.5())((

4

)()(

2

1

)()(

2

1,

2/1

21

223221

2

21

132221

21

13222122

21

mm

kkkkk

mm

mkkmkk

mm

mkkmkk

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5.3 Free-Vibration Analysis of an Undamped System

• To determine the values of X1 and X2,

• The normal modes of vibration corresponding to ω12 and ω2

2 can be

expressed, respectively, as

)11.5()(

)(

)(

)(

32222

2

2

21221

)2(1

)2(2

2

32212

2

2

21211

)1(1

)1(2

1

kkm

k

k

kkm

X

Xr

kkm

k

k

kkm

X

Xr

)12.5( and )2(

12

)2(1

)2(2

)2(1)2(

)1(11

)1(1

)1(2

)1(1)1(

Xr

X

X

XX

Xr

X

X

XX

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5.3 Free-Vibration Analysis of an Undamped System

• The free-vibration solution or the motion in time can be expressed itself as

• Initial conditions

The initial conditions are

(5.17)mode second)cos(

)cos(

)(

)()(

modefirst )cos(

)cos(

)(

)()(

22)2(

12

22)2(

1

)2(2

)2(1)2(

11)1(

11

11)1(

1

)1(2

)1(1)1(

tXr

tX

tx

txtx

tXr

tX

tx

txtx

0)0(,)0(

,0)0(constant, some )0(

2)(

12

1)(

11

txXrtx

txXtxi

i

i

Page 20: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units20

5.3 Free-Vibration Analysis of an Undamped System

• The resulting motion can be obtained by a linear superposition of the two normal modes, Eq.(5.13)

• Thus the components of the vector can be expressed as

• The unknown constants can be determined from the initial conditions:

)14.5()()()( 2211 txctxctx

)15.5()cos()cos(

)()()(

)cos()cos()()()(

22)2(

1211)1(

11

)2(2

)1(22

22)2(

111)1(

1)2(

1)1(

11

tXrtXr

txtxtx

tXtXtxtxtx

)16.5()0()0(),0()0(

),0()0(),0()0(

2222

1111

xtxxtx

xtxxtx

Page 21: Vibration Chapter05

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5.3 Free-Vibration Analysis of an Undamped System

• Substituting into Eq.(5.15) leads to

• The solution can be expressed as

)17.5(sinsin)0(

coscos)0(

sinsin)0(

coscos)0(

2)2(

1221)1(

1112

2)2(

121)1(

112

2)2(

121)1(

111

2)2(

11)1(

11

XrXrx

XrXrx

XXx

XXx

)(

)0()0(sin,

)(

)0()0(sin

)0()0(cos,

)0()0(cos

122

2112

)2(1

121

2121

)1(1

12

2112

)2(1

12

2121

)1(1

rr

xxrX

rr

xxrX

rr

xxrX

rr

xxrX

Page 22: Vibration Chapter05

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5.3 Free-Vibration Analysis of an Undamped System

• We can obtain the desired solution as

)18.5()0()0([

)0()0(tan

cos

sintan

)0()0([

)0()0(tan

cos

sintan

)0()0()0()0(

)(

1

sincos

)0()0()0()0(

)(

1

sincos

2112

2111

2)2(

1

2)2(

112

2121

2121

1)1(

1

1)1(

111

2/1

22

22112

21112

2/12

2)2(

1

2

2)2(

1)2(

1

2/1

21

22122

21212

2/12

1)1(

1

2

1)1(

1)1(

1

xxr

xxr

X

X

xxr

xxr

X

X

xxrxxr

rr

XXX

xxrxxr

rr

XXX

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© 2011 Mechanical Vibrations Fifth Edition in SI Units23

5.3 Free-Vibration Analysis of an Undamped System

Example 5.3

Free-Vibration Response of a Two-Degree-of-Freedom System

Find the free-vibration response of the system shown in Fig.5.3(a) with

k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 = c2 = c3 = 0 for the

initial conditions .

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5.3 Free-Vibration Analysis of an Undamped System

Example 5.3

Free-Vibration Response of a Two Degree of Freedom System

Solution

For the given data, the eigenvalue problem, Eq.(5.8), becomes

By setting the determinant of the coefficient matrix in Eq.(E.1) to zero, we obtain the frequency equation,

(E.1)0

0

55-

5 3510

0

0

2

1

2

2

2

1

322

22

2212

1

X

X

X

X

kkmk

kkkm

(E.2)01508510 24

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5.3 Free-Vibration Analysis of an Undamped System

Example 5.3

Free-Vibration Response of a Two-Degree-of-Freedom System

Solution

The natural frequencies can be found as

The normal modes (or eigenvectors) are given by

E.3)(4495.2,5811.1

0.6,5.2

21

22

21

E.5)(5

1

E.4)(2

1

)2(1)2(

2

)2(1)2(

)1(1)1(

2

)1(1)1(

XX

XX

XX

XX

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5.3 Free-Vibration Analysis of an Undamped System

Example 5.3

Free-Vibration Response of a Two-Degree-of-Freedom System

Solution

The free-vibration responses of the masses m1 and m2 are given by

(see Eq.5.15):

By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain

(E.7))4495.2cos(5)5811.1cos(2)(

(E.6))4495.2cos()5811.1cos()(

2)2(

11)1(

12

2)2(

11)1(

11

tXtXtx

tXtXtx

(E.11)sin2475.121622.3)0(

(E.10)sin4495.2sin5811.10)0(

(E.9)cos5cos20)0(

(E.8)coscos1)0(

2)2(

1)1(

12

2)2(

11)1(

11

2)2(

11)1(

12

2)2(

11)1(

11

XXtx

XXtx

XXtx

XXtx

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© 2011 Mechanical Vibrations Fifth Edition in SI Units27

5.3 Free-Vibration Analysis of an Undamped System

Example 5.3Free-Vibration Response of a Two-Degree-of-Freedom System Solution

The solution of Eqs.(E.8) and (E.9) yields

The solution of Eqs.(E.10) and (E.11) leads to

Equations (E.12) and (E.13) gives

(E.12)7

2cos;

7

5cos 2

)2(11

)1(1 XX

(E.13)0sin,0sin 2)2(

11)1(

1 XX

(E.14)0,0,7

2,

7

521

)2(1

)1(1 XX

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© 2011 Mechanical Vibrations Fifth Edition in SI Units28

5.3 Free-Vibration Analysis of an Undamped System

Example 5.3Free-Vibration Response of a Two-Degree-of-Freedom System Solution

Thus the free vibration responses of m1 and m2 are given by

(E.16)4495.2cos7

105811.1cos

7

10)(

(E.15)4495.2cos7

25811.1cos

7

5)(

2

1

tttx

tttx

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© 2011 Mechanical Vibrations Fifth Edition in SI Units29

5.4Torsional System

5.4

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© 2011 Mechanical Vibrations Fifth Edition in SI Units30

5.4 Torsional System

• Consider a torsional system as shown in Fig.5.6. The differential equations of rotational motion for the discs can be derived as

• Upon rearrangement become

• For the free vibration analysis of the system, Eq.(5.19) reduces to

22312222

11221111

)(

)(

ttt

ttt

MkkJ

MkkJ

)19.5()(

)(

22321222

12212111

tttt

tttt

MkkkJ

MkkkJ

)20.5(0)(

0)(

2321222

2212111

ttt

ttt

kkkJ

kkkJ

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5.4 Torsional System

Example 5.4Natural Frequencies of a Torsional System

Find the natural frequencies and mode shapes for the torsional system

shown in the figure below for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .

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© 2011 Mechanical Vibrations Fifth Edition in SI Units32

5.4 Torsional System

Example 5.4Natural Frequencies of a Torsional SystemSolution

The differential equations of motion, Eq.(5.20), reduce to (with kt3 = 0,

kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):

Rearranging and substituting the harmonic solution:

(E.1) 02

02

2120

2110

tt

tt

kkJ

kkJ

(E.2)2,1);cos()( itt ii

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© 2011 Mechanical Vibrations Fifth Edition in SI Units33

5.4 Torsional System

Example 5.4Natural Frequencies of a Torsional SystemSolution

This gives the frequency equation of

The solution of Eq.(E.3) gives the natural frequencies

(E.3)052 20

220

4 tt kkJJ

(E.4))175(4

and)175(4 0

20

1 J

k

J

k tt

Page 34: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units34

5.4 Torsional System

Example 5.4Natural Frequencies of a Torsional SystemSolution

The amplitude ratios are given by

Equations (E.4) and (E.5) can also be obtained by substituting the following in Eqs.(5.10) and (5.11).

(E.5)4

)175(2

4

)175(2

)2(1

)2(2

2

)1(1

)1(2

1

r

r

0and2,

,,

3022011

2211

kJJmJJm

kkkkkk tttt

Page 35: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units35

5.5Coordinate Coupling and Principal Coordinates

5.5

Page 36: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units36

5.5 Coordinate Coupling and Principal Coordinates

• Generalized coordinates are sets of n coordinates used to describe the configuration of the system.

• Equations of motion Using x(t) and θ(t)

Page 37: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units37

5.5 Coordinate Coupling and Principal Coordinates

From the free-body diagram shown in Figure (a), with the positive values of the motion variables as indicated, the force equilibrium equation in the vertical direction can be written as

The moment equation about C.G. can be expressed as

Eqs.(5.21) and (5.22) can be rearranged and written in matrix form as

)21.5()()( 2211 lxklxkxm

)22.5()()( 2221110 llxkllxkJ

)23.5(0

0

)( )(

)( )(

0

0 2

22

12211

221121

0 21

x

lklklklk

lklkkkx

J

m

Page 38: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units38

5.5 Coordinate Coupling and Principal Coordinates

The lathe rotates in the vertical plane and has vertical motion as

well, unless k1l1 = k2l2. This is known as elastic or static coupling.

•Equations of motion Using y(t) and θ(t)

From Figure b, the equations of motion for translation and rotation can be written as

melyklykym )()( 2211

)24.5()()( 222111 ymellykllykJP

Page 39: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units39

5.5 Coordinate Coupling and Principal Coordinates

These equations can be rearranged and written in matrix form as

If , the system will have dynamic or inertia coupling only.

•Note the following characteristics of these systems:

1.In the most general case, a viscously damped two degree of freedom system has the equations of motions in the form:

)25.5(0

0

)()(

)()(

2

22

112211

112221

2

y

lklklklk

lklkkky

Jme

mem

P

2211 lklk

)26.5(0

0

2

1

2221

1211

2

1

2221

1211

2

1

2221

1211

x

x

kk

kk

x

x

cc

cc

x

x

mm

mm

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© 2011 Mechanical Vibrations Fifth Edition in SI Units40

5.5 Coordinate Coupling and Principal Coordinates

2. The system vibrates in its own natural way regardless of the coordinates used. The choice of the coordinates is a mere convenience.

3. Principal or natural coordinates are defined as system of coordinates which give equations of motion that are uncoupled both statically and dynamically.

Page 41: Vibration Chapter05

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© 2011 Mechanical Vibrations Fifth Edition in SI Units42

5.5 Coordinate Coupling and Principal Coordinates

Example 5.6Principal Coordinates of Spring-Mass System

Determine the principal coordinates for the spring-mass system shown in the figure.

Page 43: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units43

5.5 Coordinate Coupling and Principal Coordinates

Example 5.6Principal Coordinates of Spring-Mass SystemSolution

Define two independent solutions as principal coordinates and express

them in terms of the solutions x1(t) and x2(t).

The general motion of the system shown is

(E.1)3

coscos)(

3coscos)(

22112

22111

tm

kBt

m

kBtx

tm

kBt

m

kBtx

Page 44: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units44

5.5 Coordinate Coupling and Principal Coordinates

Example 5.6Principal Coordinates of Spring-Mass SystemSolution

We define a new set of coordinates such that

Since the coordinates are harmonic functions, their corresponding equations of motion can be written as

(E.2)3

cos)(

cos)(

222

111

tm

kBtq

tm

kBtq

(E.3)03

and 0 2211

q

m

kqq

m

kq

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© 2011 Mechanical Vibrations Fifth Edition in SI Units45

5.5 Coordinate Coupling and Principal Coordinates

Example 5.6Principal Coordinates of Spring-Mass SystemSolution

From Eqs.(E.1) and (E.2), we can write

The solution of Eqs.(E.4) gives the principal coordinates:

(E.4))()()(

)()()(

212

211

tqtqtx

tqtqtx

(E.5))]()([2

1)(

)]()([2

1)(

212

211

txtxtq

txtxtq

Page 46: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units46

5.6Forced-Vibration Analysis

5.6

Page 47: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units47

5.6 Forced-Vibration Analysis

• The equations of motion of a general two-degree-of-freedom system under external forces can be written as

• Consider the external forces to be harmonic:

where ω is the forcing frequency.

• We can write the steady-state solutions as

)27.5(

2

1

2

1

2221

1211

2

1

2221

1211

2

1

2212

1211

F

F

x

x

kk

kk

x

x

cc

cc

x

x

mm

mm

)28.5(2,1,)( 0 jeFtF tijj

)29.5(2,1,)( jeXtx tijj

Page 48: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units48

5.6 Forced-Vibration Analysis

• We can write Eq.(5.30) as:

where

• Eq.(5.32) can be solved to obtain:

)32.5()( 0FXiZ

20

100

2

1

2212

1211 matrix Impedance)( )(

)( )()(

F

FF

X

XX

iZiZ

iZiZiZ

)33.5()( 01FiZX

Page 49: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units49

5.6 Forced-Vibration Analysis

• The inverse of the impedance matrix is given

• Eqs.(5.33) and (5.34) lead to the solution

)34.5()( )(

)( )(

)()()(

1)(

1112

1222

2122211

1

iZiZ

i-ZiZ

iZiZiZiZ

)35.5()()()(

)()()(

)()()(

)()()(

2122211

201110122

2122211

201210221

iZiZiZ

FiZFiZiX

iZiZiZ

FiZFiZiX

Page 50: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units50

5.6 Forced-Vibration Analysis

Example 5.8

Steady-State Response of Spring-Mass System

Find the steady-state response of system shown in Fig.5.15 when the

mass m1 is excited by the force F1(t) = F10 cos ωt. Also, plot its

frequency response curve.

Page 51: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units51

5.6 Forced-Vibration Analysis

Example 5.8

Steady-State Response of Spring-Mass System

Solution

The equations of motion of the system can be expressed as

We assume the solution to be as follows

Eq.(5.31) gives

(E.1)0

cos

2

2

0

0 10

2

1

2

1

tF

x

x

k-k

-kk

x

x

m

m

E.2)(2,1;cos)( jtXtx jj

(E.3))(,2)()( 122

2211 kZkmZZ

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© 2011 Mechanical Vibrations Fifth Edition in SI Units52

5.6 Forced-Vibration Analysis

Example 5.8

Steady-State Response of Spring-Mass System

Solution

Hence

Eqs.(E.4) and (E.5) can be expressed as

(E.5)))(3()2(

)(

(E.4)))(3(

)2(

)2(

)2()(

2210

22210

2

2210

2

22210

2

1

kmkm

kF

kkm

kFX

kmkm

Fkm

kkm

FkmX

E.6)(

1

2

)(2

1

2

1

2

1

2

10

2

1

1

k

F

X

Page 53: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units53

5.6 Forced-Vibration Analysis

Example 5.8

Steady-State Response of Spring-Mass System

Solution

E.7)(

1

)(2

1

2

1

2

1

2

102

k

FX

Page 54: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units54

5.7Semidefinite Systems

5.7

Page 55: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units55

5.7 Semidefinite Systems

• Semidefinite systems are also known as unrestrained or degenerate systems.

• Two examples of such systems are shown in the figure. For Figure (a), the equations of motion can be written as

• For free vibration, we assume the motion to be harmonic:

• Substituting Eq.(5.37) into Eq.(5.36) gives

)36.5(0)(

0)(

1222

2111

xxkxm

xxkxm

)37.5(2,1),cos()( jtXtx jjj

)38.5(0)(

0)(

22

21

212

1

XkmkX

kXXkm

Page 56: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units56

5.7 Semidefinite Systems

Page 57: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units57

5.7 Semidefinite Systems

• We obtain the frequency equation as

• From which the natural frequencies can be obtained:

• Such systems, which have one of the natural frequencies equal to zero, are called semidefinite systems.

)39.5(0)]([ 212

212 mmkmm

)40.5()(

and 021

2121 mm

mmk

Page 58: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units58

5.8Self-Excitation and Stability Analysis

5.8

Page 59: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units59

5.8 Self-Excitation and Stability Analysis

• Given that the criterion for stability is that the real parts of si must

be negative, all coefficients of equation ai must be positive and

hence, the condition

must be fulfilled.

• The Routh-Hurwitz criterion states that the system will be stable if

all the coefficients a0, a1,…,a4 are positive and the determinants

defined below are positive:

)46.5(214

230321 aaaaaaa

Page 60: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units60

5.8 Self-Excitation and Stability Analysis

)49.5(0

0

0

)48.5(0

)47.5(0

2304

21321

31

420

31

3

302120

312

11

aaaaaaa

aa

aaa

aa

T

aaaaaa

aaT

aT

Page 61: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units61

5.9Transfer-Function Approach

5.9

Page 62: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units62

5.9 Transfer-Function Approach

• For two-degree-of-freedom system shown the equations of motion are

• By taking the Laplace transforms of Eqs. (5.50) and (5.51), assuming zero initial conditions,

5.51

5.50

2122321223222

1221212212111

fxkxkkxcxccxm

fxkxkkxcxccxm

5.53

5.52

2122321223222

2

1221212212112

1

sFsXksXkkssXcssXccsXsm

sFsXksXkkssXcssXccsXsm

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© 2011 Mechanical Vibrations Fifth Edition in SI Units63

5.9 Transfer-Function Approach

• It can be solved using Cramer’s rule as

where

5.57 and 5.56 22

11 sD

sDsX

sD

sDsX

Page 64: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units64

5.9 Transfer-Function Approach

• Note that

1. The denominator, D(s), in the expressions of X1(s) and X2(s)

given by Eq. (5.60), is a fourth-order polynomial in s and denotes the characteristic polynomial of the system The model (or system) is a fourth-order model (or system).

2. Equations (5.56) and (5.57) permit us to apply inverse Laplace transforms to obtain the fourth-order differential equations for

x1(t) and x2(t) .

3. Equations (5.56) and (5.57) can be used to derive the transfer

functions of and x2(t) corresponding to any specified forcing

function.

Page 65: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units65

5.10Solutions Using Laplace Transform

5.10

Page 66: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units66

5.10 Solutions Using Laplace Transform

Example 5.12

Response Under Impulse Using Laplace Transform Method

Two railway cars, of masses m1 = M and m2 = m are connected by a

spring of stiffness k, as shown in the figure. If the car of mass M is subjected to an impulse determine the time responses of the cars using the Laplace transform method.

tF 0

Page 67: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units67

5.10 Solutions Using Laplace Transform

Example 5.12

Response Under Impulse Using Laplace Transform Method

Solution

The responses of the cars can be determined using either of the following approaches:a. Consider the system to be undergoing free vibration due to the initial velocity caused by the impulse applied to car M.b. Consider the system to be undergoing forced vibration due to the force applied to car M (with the displacements and velocities of cars M and m considered to be zero initially).

tF 0

Page 68: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units68

5.10 Solutions Using Laplace Transform

Example 5.12

Response Under Impulse Using Laplace Transform Method

Solution

Using the second approach, the equations of motion of the cars can be expressed as

Using the Laplace transforms, Eqs. (E.1) and (E.2) can be written as

E.2 0

E.1

122

0211

xxkxm

tFxxkxM

E.4 0

E.3

22

1

0212

sXkmsskX

FskXsXkMs

Page 69: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units69

5.10 Solutions Using Laplace Transform

Example 5.12

Response Under Impulse Using Laplace Transform Method

Solution

Equations (E.3) and (E.4) can be solved as

E.6

E.5

220

2

22

20

1

mMkMmss

kFsX

mMkMmss

kmsFsX

Page 70: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units70

5.10 Solutions Using Laplace Transform

Example 5.12

Response Under Impulse Using Laplace Transform Method

Solution

Using partial fractions, Eqs. (E.5) and (E.6) can be rewritten as

where

E.8 11

E.7 1

2220

2

2220

1

ws

w

wsmM

FsX

ws

w

wM

m

smM

FsX

E.9 112

mMkw

Page 71: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units71

5.10 Solutions Using Laplace Transform

Example 5.12

Response Under Impulse Using Laplace Transform Method

Solution

The inverse transforms of Eqs. (E.7) and (E.8), using the results of Appendix D, yield the time responses of the cars as

E.11 sin1

E.10 sin

02

01

wtw

tmM

FsX

wtwM

mt

mM

Fsx

Page 72: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units72

5.11Solutions Using Frequency Transfer Functions

5.11

Page 73: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units73

5.11 Solutions Using Frequency Transfer Functions

Example 5.13

Derivation of Frequency Transfer Functions

Derive the frequency transfer functions of x1(t) and x2(t) for the system

shown in figure.

Page 74: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units74

5.11 Solutions Using Frequency Transfer Functions

Example 5.13

Derivation of Frequency Transfer Functions

Solution

From the free-body diagrams of the masses, the equations of motion of the system is

E.2 0

E.1 sin

212212222

01212212111111

pxxkxxcxm

wtPpxxkxxcxkxcxm

Page 75: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units75

5.11 Solutions Using Frequency Transfer Functions

Example 5.13

Derivation of Frequency Transfer Functions

Solution

By taking the Laplace transforms of Eqs. (E.1) and (E.2), assuming zero initial conditions,

E.4 0

E.3

12212222

1212212111112

1

sXsXksXsXcsXm

sPsXsXksXsXcsXkssXcsXsm

Page 76: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units76

5.11 Solutions Using Frequency Transfer Functions

Example 5.13

Derivation of Frequency Transfer Functions

Solution

The of Eqs. (E.3) and (E.4) is

where

E.6 and E.5 22

11 sD

sDsX

sD

sDsX

E.7

E.7

1222

1222

21

sPkscsD

sPkscsmsD

Page 77: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units77

5.11 Solutions Using Frequency Transfer Functions

Example 5.13

Derivation of Frequency Transfer Functions

Solution

We have

The general transfer functions is

E.9 211221

221221221

3221221

422

kkskckc

scckmkmkmscmcmcmsmmsD

E.10 and E.9 22

1

2222

2

1

1

sD

ksc

sP

sX

sD

kscsm

sP

sX

Page 78: Vibration Chapter05

© 2011 Mechanical Vibrations Fifth Edition in SI Units78

5.11 Solutions Using Frequency Transfer Functions

Example 5.13

Derivation of Frequency Transfer Functions

Solution

The frequency transfer functions is

where

E.13 and E.12 22

1

2222

2

1

1

iwD

kiwc

iwP

iwX

iwD

kiwcwm

iwP

iwX

21122121221221

2

2212214

214

kkkckciwcckmkmkmw

cmcmcmiwwmmwiwD